Theory of EquationsmediumPYQ · JEE Main · 6 Apr 2026 · Shift 1 (Morning)Free

Eccentricities from a Quadratic: α²+β²+γ² = 26 | JEE Main 2026

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Question
Let one root of the quadratic equation (k2−15k+27)x2+9(k−1)x+18=0(k^2-15k+27)x^2+9(k-1)x+18=0 be twice the other. Then the length of the latus rectum of the parabola y2=6kxy^2=6kx is equal to
A44
B66
C88
D1212correct
Solution
Step 1: Let A=k2−15k+27A=k^2-15k+27 and roots α,2α\alpha,2\alpha.
α+2α=3α=−9(k−1)A,α⋅2α=2α2=18A.\alpha+2\alpha=3\alpha=\dfrac{-9(k-1)}{A},\qquad\alpha\cdot2\alpha=2\alpha^2=\dfrac{18}{A}.
 Step 2: α=−3(k−1)A⇒2⋅9(k−1)2A2=18A⇒18(k−1)2A2=18A\alpha=\dfrac{-3(k-1)}{A}\Rightarrow2\cdot\dfrac{9(k-1)^2}{A^2}=\dfrac{18}{A}\Rightarrow\dfrac{18(k-1)^2}{A^2}=\dfrac{18}{A}.
⇒(k−1)2=A=k2−15k+27⇒k2−2k+1=k2−15k+27.\Rightarrow(k-1)^2=A=k^2-15k+27\Rightarrow k^2-2k+1=k^2-15k+27.
⇒−2k+1=−15k+27⇒13k=26⇒k=2.\Rightarrow-2k+1=-15k+27\Rightarrow13k=26\Rightarrow k=2.
 Step 3: y2=6kx=6(2)x=12x=4(3)xy^2=6kx=6(2)x=12x=4(3)x.
∴latus rectum=4(3)=12.\therefore\text{latus rectum}=4(3)=12.
 Correct answer: (4)
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